Solve by Factoring z/(z-5)+3/(z+5)=50/(z^2-25)

$\frac{z}{z - 5} + \frac{3}{z + 5} = \frac{50}{z^{2} - 25}$

Step 1

Subtract $\frac{50}{z^{2} - 25}$ from both sides of the equation.

$\frac{z}{z - 5} + \frac{3}{z + 5} - \frac{50}{z^{2} - 25} = 0$

Step 2

Simplify $\frac{z}{z - 5} + \frac{3}{z + 5} - \frac{50}{z^{2} - 25}$ .

Simplify the denominator.

Rewrite $25$ as $5^{2}$ .

$\frac{z}{z - 5} + \frac{3}{z + 5} - \frac{50}{z^{2} - 5^{2}} = 0$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = z$ and $b = 5$ .

$\frac{z}{z - 5} + \frac{3}{z + 5} - \frac{50}{(z + 5) (z - 5)} = 0$

To write $\frac{z}{z - 5}$ as a fraction with a common denominator, multiply by $\frac{z + 5}{z + 5}$ .

$\frac{z}{z - 5} \cdot \frac{z + 5}{z + 5} + \frac{3}{z + 5} - \frac{50}{(z + 5) (z - 5)} = 0$

To write $\frac{3}{z + 5}$ as a fraction with a common denominator, multiply by $\frac{z - 5}{z - 5}$ .

$\frac{z}{z - 5} \cdot \frac{z + 5}{z + 5} + \frac{3}{z + 5} \cdot \frac{z - 5}{z - 5} - \frac{50}{(z + 5) (z - 5)} = 0$

Write each expression with a common denominator of $(z - 5) (z + 5)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{z}{z - 5}$ by $\frac{z + 5}{z + 5}$ .

$\frac{z (z + 5)}{(z - 5) (z + 5)} + \frac{3}{z + 5} \cdot \frac{z - 5}{z - 5} - \frac{50}{(z + 5) (z - 5)} = 0$

Multiply $\frac{3}{z + 5}$ by $\frac{z - 5}{z - 5}$ .

$\frac{z (z + 5)}{(z - 5) (z + 5)} + \frac{3 (z - 5)}{(z + 5) (z - 5)} - \frac{50}{(z + 5) (z - 5)} = 0$

Reorder the factors of $(z - 5) (z + 5)$ .

$\frac{z (z + 5)}{(z + 5) (z - 5)} + \frac{3 (z - 5)}{(z + 5) (z - 5)} - \frac{50}{(z + 5) (z - 5)} = 0$

Combine the numerators over the common denominator.

$\frac{z (z + 5) + 3 (z - 5)}{(z + 5) (z - 5)} - \frac{50}{(z + 5) (z - 5)} = 0$

Combine the numerators over the common denominator.

$\frac{z (z + 5) + 3 (z - 5) - 50}{(z + 5) (z - 5)} = 0$

Simplify each term.

Apply the distributive property.

$\frac{z \cdot z + z \cdot 5 + 3 (z - 5) - 50}{(z + 5) (z - 5)} = 0$

Multiply $z$ by $z$ .

$\frac{z^{2} + z \cdot 5 + 3 (z - 5) - 50}{(z + 5) (z - 5)} = 0$

Move $5$ to the left of $z$ .

$\frac{z^{2} + 5 \cdot z + 3 (z - 5) - 50}{(z + 5) (z - 5)} = 0$

Apply the distributive property.

$\frac{z^{2} + 5 z + 3 z + 3 \cdot - 5 - 50}{(z + 5) (z - 5)} = 0$

Multiply $3$ by $- 5$ .

$\frac{z^{2} + 5 z + 3 z - 15 - 50}{(z + 5) (z - 5)} = 0$

Add $5 z$ and $3 z$ .

$\frac{z^{2} + 8 z - 15 - 50}{(z + 5) (z - 5)} = 0$

Subtract $50$ from $- 15$ .

$\frac{z^{2} + 8 z - 65}{(z + 5) (z - 5)} = 0$

Factor $z^{2} + 8 z - 65$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 65$ and whose sum is $8$ .

$- 5, 13$

Write the factored form using these integers.

$\frac{(z - 5) (z + 13)}{(z + 5) (z - 5)} = 0$

Cancel the common factor of $z - 5$ .

Cancel the common factor.

$\frac{(z - 5) (z + 13)}{(z + 5) (z - 5)} = 0$

Rewrite the expression.

$\frac{z + 13}{z + 5} = 0$

Step 3

Set the numerator equal to zero.

$z + 13 = 0$

Step 4

Subtract $13$ from both sides of the equation.

$z = - 13$

Solve by Factoring z/(z-5)+3/(z+5)=50/(z^2-25)

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